# Am I doing this right?

• Feb 18th 2010, 11:45 AM
Tall Jessica
Am I doing this right?
It's a simplification. I'm supposed to simplify so that only one variable is present at the end.

x = [sin(A+B)/2]/sin(A/2)

Here's what I've done so far...

x = 2[sin(A+B)] / 2[sin(A)

x = sin(A + B) / sin(A)

am I totally off....?

Thank you!
• Feb 18th 2010, 04:37 PM
skeeter
Quote:

Originally Posted by Tall Jessica
It's a simplification. I'm supposed to simplify so that only one variable is present at the end.

x = [sin(A+B)/2]/sin(A/2)

Here's what I've done so far...

x = 2[sin(A+B)] / 2[sin(A)

x = sin(A + B) / sin(A)

am I totally off....?

Thank you!

is the equation ...

$\displaystyle x = \frac{\sin\left(\frac{A+B}{2}\right)}{\sin\left(\f rac{A}{2}\right)}$

or

$\displaystyle x = \frac{\frac{\sin(A+B)}{2}}{\sin\left(\frac{A}{2}\r ight)}$

or something else ?
• Feb 18th 2010, 05:07 PM
Tall Jessica
Quote:

Originally Posted by skeeter
is the equation ...

$\displaystyle x = \frac{\sin\left(\frac{A+B}{2}\right)}{\sin\left(\f rac{A}{2}\right)}$

or

$\displaystyle x = \frac{\frac{\sin(A+B)}{2}}{\sin\left(\frac{A}{2}\r ight)}$

or something else ?

The first one. Thanks :-)
• Feb 18th 2010, 05:51 PM
skeeter
I don't know if the original expression can be "simplified" any more than it is already. You're not going to get an expression with only A or B alone unless there is a known relationship between A and B.

I tried using the half-angle formula for sine ...

$\displaystyle \sin\left(\frac{A+B}{2}\right) = \pm \sqrt{\frac{1 - \cos(A+B)}{2}}$

$\displaystyle \sin\left(\frac{A}{2}\right) = \pm \sqrt{\frac{1 - \cos{A}}{2}}$

doing the division ...

$\displaystyle \pm \sqrt{\frac{1 - \cos(A+B)}{1 - \cos{A}}}$

I also did the sum expansion for $\displaystyle \sin\left(\frac{A}{2} + \frac{B}{2}\right)$ , divided by $\displaystyle \sin\left(\frac{A}{2}\right)$ and got the following expression ...

$\displaystyle \cos\left(\frac{B}{2}\right) + \cot\left(\frac{A}{2}\right) \sin\left(\frac{B}{2}\right)$

not much there as far as "simplification" goes, either.

Is there more to this problem that you haven't posted?
• Feb 18th 2010, 06:20 PM
Tall Jessica
Well... I probably copied it wrong. I'll check tomorrow.

Thank you and sorry!